§ 4.6 Applied Optimization

Practical Tips

1 Make sure you know what quantity is being optimized.

2 Sketch a diagram of the situation and label all variables in thesketch.

3 Obtain a formula for the quantity to be optimized and eliminateall variables but one (where applicable). Find the domain.

4 Find the critical points and evaluate at these points and theendpoints to find the global minimum or maximum, whichever isneeded.

Practical Tips

1 Make sure you know what quantity is being optimized.2 Sketch a diagram of the situation and label all variables in the

sketch.

3 Obtain a formula for the quantity to be optimized and eliminateall variables but one (where applicable). Find the domain.

4 Find the critical points and evaluate at these points and theendpoints to find the global minimum or maximum, whichever isneeded.

Practical Tips

1 Make sure you know what quantity is being optimized.2 Sketch a diagram of the situation and label all variables in the

sketch.3 Obtain a formula for the quantity to be optimized and eliminate

all variables but one (where applicable). Find the domain.

4 Find the critical points and evaluate at these points and theendpoints to find the global minimum or maximum, whichever isneeded.

Practical Tips

1 Make sure you know what quantity is being optimized.2 Sketch a diagram of the situation and label all variables in the

sketch.3 Obtain a formula for the quantity to be optimized and eliminate

all variables but one (where applicable). Find the domain.4 Find the critical points and evaluate at these points and the

endpoints to find the global minimum or maximum, whichever isneeded.

Fencing Example

Example

You want to fence a rectangular region of area 1000 ft2. You choosetwo different kinds of fencing to use: the parallel sides on the frontand back of the region will be fenced with $5 per foot fencing. Theleft and right sides will cost $3 per foot. Minimize the cost.

What quantity are we trying to optimize?

Is this a minimization or maximization problem?

What is our domain? 0 < x <∞

Fencing Example

Example

You want to fence a rectangular region of area 1000 ft2. You choosetwo different kinds of fencing to use: the parallel sides on the frontand back of the region will be fenced with $5 per foot fencing. Theleft and right sides will cost $3 per foot. Minimize the cost.

What quantity are we trying to optimize?

Is this a minimization or maximization problem?

What is our domain? 0 < x <∞

Fencing Example

Example

You want to fence a rectangular region of area 1000 ft2. You choosetwo different kinds of fencing to use: the parallel sides on the frontand back of the region will be fenced with $5 per foot fencing. Theleft and right sides will cost $3 per foot. Minimize the cost.

What quantity are we trying to optimize?

Is this a minimization or maximization problem?

What is our domain? 0 < x <∞

Fencing Example

Example

You want to fence a rectangular region of area 1000 ft2. You choosetwo different kinds of fencing to use: the parallel sides on the frontand back of the region will be fenced with $5 per foot fencing. Theleft and right sides will cost $3 per foot. Minimize the cost.

What quantity are we trying to optimize?

Is this a minimization or maximization problem?

What is our domain?

0 < x <∞

Fencing Example

Example

You want to fence a rectangular region of area 1000 ft2. You choosetwo different kinds of fencing to use: the parallel sides on the frontand back of the region will be fenced with $5 per foot fencing. Theleft and right sides will cost $3 per foot. Minimize the cost.

What quantity are we trying to optimize?

Is this a minimization or maximization problem?

What is our domain? 0 < x <∞

Fencing Example

Now we need to visualize the situation.

5x

3yA = xy = 1000 ft2

C=10x+6y

Fencing Example

Now we need to visualize the situation.

5x

3yA = xy = 1000 ft2

C=10x+6y

Fencing Example

Now we need to visualize the situation.

5x

3yA = xy = 1000 ft2

C=10x+6y

Fencing Example

Now we need to visualize the situation.

5x

3y

A = xy = 1000 ft2

C=10x+6y

Fencing Example

Now we need to visualize the situation.

5x

3yA = xy = 1000 ft2

C=10x+6y

Fencing Example

Now we need to visualize the situation.

5x

3yA = xy = 1000 ft2

C=10x+6y

Fencing Example

We have two variables, which is why we need the two equations tosolve. What can we do?

We need to rewrite one equation to be able to write the other equationin terms of only one variable. Which way should we approach this?

A = xy = 1000

C = 10x + 6y

xy = 1000

y =1000

x

Fencing Example

We have two variables, which is why we need the two equations tosolve. What can we do?

We need to rewrite one equation to be able to write the other equationin terms of only one variable. Which way should we approach this?

A = xy = 1000

C = 10x + 6y

xy = 1000

y =1000

x

Fencing Example

We have two variables, which is why we need the two equations tosolve. What can we do?

We need to rewrite one equation to be able to write the other equationin terms of only one variable. Which way should we approach this?

A = xy = 1000

C = 10x + 6y

xy = 1000

y =1000

x

Fencing Example

We have two variables, which is why we need the two equations tosolve. What can we do?

We need to rewrite one equation to be able to write the other equationin terms of only one variable. Which way should we approach this?

A = xy = 1000

C = 10x + 6y

xy = 1000

y =1000

x

Fencing Example

C = 10x + 6y

C = 10x +6000

x

C′ = 10− 6000x2 = 0

10 =6000

x2

x2 = 600

x =√

600 ≈ 24.49 ft

Fencing Example

C = 10x + 6y

C = 10x +6000

x

C′ = 10− 6000x2 = 0

10 =6000

x2

x2 = 600

x =√

600 ≈ 24.49 ft

Fencing Example

C = 10x + 6y

C = 10x +6000

x

C′ =

10− 6000x2 = 0

10 =6000

x2

x2 = 600

x =√

600 ≈ 24.49 ft

Fencing Example

C = 10x + 6y

C = 10x +6000

x

C′ = 10− 6000x2

= 0

10 =6000

x2

x2 = 600

x =√

600 ≈ 24.49 ft

Fencing Example

C = 10x + 6y

C = 10x +6000

x

C′ = 10− 6000x2 = 0

10 =6000

x2

x2 = 600

x =√

600 ≈ 24.49 ft

Fencing Example

C = 10x + 6y

C = 10x +6000

x

C′ = 10− 6000x2 = 0

10 =6000

x2

x2 = 600

x =√

600 ≈ 24.49 ft

Fencing Example

C = 10x + 6y

C = 10x +6000

x

C′ = 10− 6000x2 = 0

10 =6000

x2

x2 = 600

x =√

600 ≈ 24.49 ft

Fencing Example

C = 10x + 6y

C = 10x +6000

x

C′ = 10− 6000x2 = 0

10 =6000

x2

x2 = 600

x =√

600 ≈ 24.49

ft

Fencing Example

C = 10x + 6y

C = 10x +6000

x

C′ = 10− 6000x2 = 0

10 =6000

x2

x2 = 600

x =√

600 ≈ 24.49 ft

Fencing Example

Does this give us a minimum cost? How can we figure that out?

C′ = 10− 6000x2

C′′ =12000

x3

C′′∣∣x=24.49 > 0

So, we have a local minimum. Is this the only value we need toconsider?

Fencing Example

Does this give us a minimum cost? How can we figure that out?

C′ = 10− 6000x2

C′′ =12000

x3

C′′∣∣x=24.49 > 0

So, we have a local minimum. Is this the only value we need toconsider?

Fencing Example

Does this give us a minimum cost? How can we figure that out?

C′ = 10− 6000x2

C′′ =12000

x3

C′′∣∣x=24.49 > 0

So, we have a local minimum. Is this the only value we need toconsider?

Fencing Example

Does this give us a minimum cost? How can we figure that out?

C′ = 10− 6000x2

C′′ =12000

x3

C′′∣∣x=24.49 > 0

So, we have a local minimum. Is this the only value we need toconsider?

Fencing Example

Does this give us a minimum cost? How can we figure that out?

C′ = 10− 6000x2

C′′ =12000

x3

C′′∣∣x=24.49 > 0

So, we have a local minimum. Is this the only value we need toconsider?

Fencing Example

Is this x value what we need?

We need the cost ...

C = 10(24.49) +600024.49

= $489.90

Fencing Example

Is this x value what we need?

We need the cost ...

C = 10(24.49) +600024.49

= $489.90

Fencing Example

Is this x value what we need?

We need the cost ...

C = 10(24.49) +600024.49

= $489.90

Fencing Example

Is this x value what we need?

We need the cost ...

C = 10(24.49) +600024.49

= $489.90

Circular Cone Example

ExampleFind the volume of the largest right circular cone that can be inscribedin a sphere of radius 19.

What quantity are we trying to optimize?

We want to maximize the volume of the cone. What is the volume ofa right circular cone?

The volume of a right circular cone is 13πr2h where r = 19 units.

What is the domain for the radius? 0 ≤ r ≤ 19

Circular Cone Example

ExampleFind the volume of the largest right circular cone that can be inscribedin a sphere of radius 19.

What quantity are we trying to optimize?

We want to maximize the volume of the cone. What is the volume ofa right circular cone?

The volume of a right circular cone is 13πr2h where r = 19 units.

What is the domain for the radius? 0 ≤ r ≤ 19

Circular Cone Example

ExampleFind the volume of the largest right circular cone that can be inscribedin a sphere of radius 19.

What quantity are we trying to optimize?

We want to maximize the volume of the cone. What is the volume ofa right circular cone?

The volume of a right circular cone is 13πr2h where r = 19 units.

What is the domain for the radius? 0 ≤ r ≤ 19

Circular Cone Example

ExampleFind the volume of the largest right circular cone that can be inscribedin a sphere of radius 19.

What quantity are we trying to optimize?

We want to maximize the volume of the cone. What is the volume ofa right circular cone?

The volume of a right circular cone is 13πr2h where r = 19 units.

What is the domain for the radius? 0 ≤ r ≤ 19

Circular Cone Example

ExampleFind the volume of the largest right circular cone that can be inscribedin a sphere of radius 19.

What quantity are we trying to optimize?

We want to maximize the volume of the cone. What is the volume ofa right circular cone?

The volume of a right circular cone is 13πr2h where r = 19 units.

What is the domain for the radius?

0 ≤ r ≤ 19

Circular Cone Example

ExampleFind the volume of the largest right circular cone that can be inscribedin a sphere of radius 19.

What quantity are we trying to optimize?

We want to maximize the volume of the cone. What is the volume ofa right circular cone?

The volume of a right circular cone is 13πr2h where r = 19 units.

What is the domain for the radius? 0 ≤ r ≤ 19

Circular Cone Example

We need to now visualize the situation.

Circular Cone Example

We need to now visualize the situation.

Circular Cone Example

Let s be the radius of the cone. We have that r is the hypotenuse of theright triangle. What equation can we write to relate the value of x tothe rest of the problem?

x = h− 19 .

Now? By the Pythagorean Theorem,

s2 = 192 − x2 = 361− x2

So, the radius is a function of x. We want the same for the height ofthe cone because we need one variable. What is this equation?h = 19 + x .

Circular Cone Example

Let s be the radius of the cone. We have that r is the hypotenuse of theright triangle. What equation can we write to relate the value of x tothe rest of the problem? x = h− 19 .

Now? By the Pythagorean Theorem,

s2 = 192 − x2 = 361− x2

So, the radius is a function of x. We want the same for the height ofthe cone because we need one variable. What is this equation?h = 19 + x .

Circular Cone Example

Let s be the radius of the cone. We have that r is the hypotenuse of theright triangle. What equation can we write to relate the value of x tothe rest of the problem? x = h− 19 .

Now?

By the Pythagorean Theorem,

s2 = 192 − x2 = 361− x2

So, the radius is a function of x. We want the same for the height ofthe cone because we need one variable. What is this equation?h = 19 + x .

Circular Cone Example

Let s be the radius of the cone. We have that r is the hypotenuse of theright triangle. What equation can we write to relate the value of x tothe rest of the problem? x = h− 19 .

Now? By the Pythagorean Theorem,

s2 = 192 − x2 = 361− x2

So, the radius is a function of x. We want the same for the height ofthe cone because we need one variable. What is this equation?h = 19 + x .

Circular Cone Example

Let s be the radius of the cone. We have that r is the hypotenuse of theright triangle. What equation can we write to relate the value of x tothe rest of the problem? x = h− 19 .

Now? By the Pythagorean Theorem,

s2 = 192 − x2 = 361− x2

So, the radius is a function of x. We want the same for the height ofthe cone because we need one variable. What is this equation?

h = 19 + x .

Circular Cone Example

Let s be the radius of the cone. We have that r is the hypotenuse of theright triangle. What equation can we write to relate the value of x tothe rest of the problem? x = h− 19 .

Now? By the Pythagorean Theorem,

s2 = 192 − x2 = 361− x2

So, the radius is a function of x. We want the same for the height ofthe cone because we need one variable. What is this equation?h = 19 + x .

Circular Cone Example

Now,

V =13πr2h

=π

3(361− x2)(19 + x)

=π

3(193 + 192x− 19x2 − x3)

Now what?

V ′ =π

3(192 − 38x− 3x2) = 0

π

3(19 + x)(19− 3x) = 0

x = −19,193

The negative root is impossible, so we have only one option.

Circular Cone Example

Now,

V =13πr2h

=π

3(361− x2)(19 + x)

=π

3(193 + 192x− 19x2 − x3)

Now what?

V ′ =π

3(192 − 38x− 3x2) = 0

π

3(19 + x)(19− 3x) = 0

x = −19,193

The negative root is impossible, so we have only one option.

Circular Cone Example

Now,

V =13πr2h

=π

3(361− x2)(19 + x)

=π

3(193 + 192x− 19x2 − x3)

Now what?

V ′ =π

3(192 − 38x− 3x2) = 0

π

3(19 + x)(19− 3x) = 0

x = −19,193

The negative root is impossible, so we have only one option.

Circular Cone Example

Now,

V =13πr2h

=π

3(361− x2)(19 + x)

=π

3(193 + 192x− 19x2 − x3)

Now what?

V ′ =π

3(192 − 38x− 3x2) = 0

π

3(19 + x)(19− 3x) = 0

x = −19,193

The negative root is impossible, so we have only one option.

Circular Cone Example

Now,

V =13πr2h

=π

3(361− x2)(19 + x)

=π

3(193 + 192x− 19x2 − x3)

Now what?

V ′ =π

3(192 − 38x− 3x2)

= 0π

3(19 + x)(19− 3x) = 0

x = −19,193

The negative root is impossible, so we have only one option.

Circular Cone Example

Now,

V =13πr2h

=π

3(361− x2)(19 + x)

=π

3(193 + 192x− 19x2 − x3)

Now what?

V ′ =π

3(192 − 38x− 3x2) = 0

π

3(19 + x)(19− 3x) = 0

x = −19,193

The negative root is impossible, so we have only one option.

Circular Cone Example

Now,

V =13πr2h

=π

3(361− x2)(19 + x)

=π

3(193 + 192x− 19x2 − x3)

Now what?

V ′ =π

3(192 − 38x− 3x2) = 0

π

3(19 + x)(19− 3x) = 0

x = −19,193

The negative root is impossible, so we have only one option.

Circular Cone Example

Now,

V =13πr2h

=π

3(361− x2)(19 + x)

=π

3(193 + 192x− 19x2 − x3)

Now what?

V ′ =π

3(192 − 38x− 3x2) = 0

π

3(19 + x)(19− 3x) = 0

x = −19,193

The negative root is impossible, so we have only one option.

Circular Cone Example

Now,

V =13πr2h

=π

3(361− x2)(19 + x)

=π

3(193 + 192x− 19x2 − x3)

Now what?

V ′ =π

3(192 − 38x− 3x2) = 0

π

3(19 + x)(19− 3x) = 0

x = −19,193

The negative root is impossible, so we have only one option.

Circular Cone Example

What do we do with this?

V ′′ = − 38− 6x

V ′′∣∣x= 19

3< 0

Conclusion? r = 193 gives a local maximum.

Circular Cone Example

What do we do with this?

V ′′ =

− 38− 6x

V ′′∣∣x= 19

3< 0

Conclusion? r = 193 gives a local maximum.

Circular Cone Example

What do we do with this?

V ′′ = − 38− 6x

V ′′∣∣x= 19

3< 0

Conclusion? r = 193 gives a local maximum.

Circular Cone Example

What do we do with this?

V ′′ = − 38− 6x

V ′′∣∣x= 19

3

< 0

Conclusion? r = 193 gives a local maximum.

Circular Cone Example

What do we do with this?

V ′′ = − 38− 6x

V ′′∣∣x= 19

3< 0

Conclusion? r = 193 gives a local maximum.

Circular Cone Example

What do we do with this?

V ′′ = − 38− 6x

V ′′∣∣x= 19

3< 0

Conclusion?

r = 193 gives a local maximum.

Circular Cone Example

What do we do with this?

V ′′ = − 38− 6x

V ′′∣∣x= 19

3< 0

Conclusion? r = 193 gives a local maximum.

Circular Cone Example

So is r = 193 units our answer?

V =π

3

(361− 361

9

)(19 +

193

)=

21948881

π ≈ 8512.86 units3.

Circular Cone Example

So is r = 193 units our answer?

V =π

3

(361− 361

9

)(19 +

193

)=

21948881

π ≈ 8512.86 units3.

Circular Cone Example

So is r = 193 units our answer?

V =π

3

(361− 361

9

)(19 +

193

)=

21948881

π ≈ 8512.86 units3.

Circular Cylinder Example

ExampleFind the radius and height of the right circular cylinder of largestvolume that can be inscribed in a right circular cone with radius 6′′

and height 10′′.

What is the quantity we are optimizing?

We want to find the dimensions that maximize the volume of a rightcircular cylinder.

What geometric formulas do we need here?

Right circular cone: Vco = 13πr2h

Right circular cylinder: Vcy = πr2h

Circular Cylinder Example

ExampleFind the radius and height of the right circular cylinder of largestvolume that can be inscribed in a right circular cone with radius 6′′

and height 10′′.

What is the quantity we are optimizing?

We want to find the dimensions that maximize the volume of a rightcircular cylinder.

What geometric formulas do we need here?

Right circular cone: Vco = 13πr2h

Right circular cylinder: Vcy = πr2h

Circular Cylinder Example

ExampleFind the radius and height of the right circular cylinder of largestvolume that can be inscribed in a right circular cone with radius 6′′

and height 10′′.

What is the quantity we are optimizing?

We want to find the dimensions that maximize the volume of a rightcircular cylinder.

What geometric formulas do we need here?

Right circular cone: Vco = 13πr2h

Right circular cylinder: Vcy = πr2h

Circular Cylinder Example

ExampleFind the radius and height of the right circular cylinder of largestvolume that can be inscribed in a right circular cone with radius 6′′

and height 10′′.

What is the quantity we are optimizing?

We want to find the dimensions that maximize the volume of a rightcircular cylinder.

What geometric formulas do we need here?

Right circular cone: Vco = 13πr2h

Right circular cylinder: Vcy = πr2h

Circular Cylinder Example

ExampleFind the radius and height of the right circular cylinder of largestvolume that can be inscribed in a right circular cone with radius 6′′

and height 10′′.

What is the quantity we are optimizing?

We want to find the dimensions that maximize the volume of a rightcircular cylinder.

What geometric formulas do we need here?

Right circular cone:

Vco = 13πr2h

Right circular cylinder: Vcy = πr2h

Circular Cylinder Example

ExampleFind the radius and height of the right circular cylinder of largestvolume that can be inscribed in a right circular cone with radius 6′′

and height 10′′.

What is the quantity we are optimizing?

We want to find the dimensions that maximize the volume of a rightcircular cylinder.

What geometric formulas do we need here?

Right circular cone: Vco = 13πr2h

Right circular cylinder: Vcy = πr2h

Circular Cylinder Example

ExampleFind the radius and height of the right circular cylinder of largestvolume that can be inscribed in a right circular cone with radius 6′′

and height 10′′.

What is the quantity we are optimizing?

We want to find the dimensions that maximize the volume of a rightcircular cylinder.

What geometric formulas do we need here?

Right circular cone: Vco = 13πr2h

Right circular cylinder:

Vcy = πr2h

Circular Cylinder Example

ExampleFind the radius and height of the right circular cylinder of largestvolume that can be inscribed in a right circular cone with radius 6′′

and height 10′′.

What is the quantity we are optimizing?

We want to find the dimensions that maximize the volume of a rightcircular cylinder.

What geometric formulas do we need here?

Right circular cone: Vco = 13πr2h

Right circular cylinder: Vcy = πr2h

Circular Cylinder Example

What visual do we need here?

Circular Cylinder Example

What visual do we need here?

Circular Cylinder Example

Since there are two variables here (r and h) we need to find anotherrelationship between them. To do this we will use a cross- section.

10

6

r

h

10− h 106 = 10−h

r

h = − 53 r + 10

Circular Cylinder Example

Since there are two variables here (r and h) we need to find anotherrelationship between them. To do this we will use a cross- section.

10

6

r

h

10− h 106 = 10−h

r

h = − 53 r + 10

Circular Cylinder Example

Since there are two variables here (r and h) we need to find anotherrelationship between them. To do this we will use a cross- section.

10

6

r

h

10− h 106 = 10−h

r

h = − 53 r + 10

Circular Cylinder Example

Since there are two variables here (r and h) we need to find anotherrelationship between them. To do this we will use a cross- section.

10

6

r

h

10− h 106 = 10−h

r

h = − 53 r + 10

Circular Cylinder Example

Since there are two variables here (r and h) we need to find anotherrelationship between them. To do this we will use a cross- section.

10

6

r

h

10− h

106 = 10−h

r

h = − 53 r + 10

Circular Cylinder Example

Since there are two variables here (r and h) we need to find anotherrelationship between them. To do this we will use a cross- section.

10

6

r

h

10− h 106 = 10−h

r

h = − 53 r + 10

Circular Cylinder Example

Since there are two variables here (r and h) we need to find anotherrelationship between them. To do this we will use a cross- section.

10

6

r

h

10− h 106 = 10−h

r

h = − 53 r + 10

Circular Cylinder Example

What is our domain for the radius here?

0 < r < 6

V = πr2(

10− 53

r)

= 10πr2 − 5π3

r3

dVdr

= 20πr − 5πr2 = 0

5πr(4− r) = 0

r = 0, 4

Circular Cylinder Example

What is our domain for the radius here? 0 < r < 6

V = πr2(

10− 53

r)

= 10πr2 − 5π3

r3

dVdr

= 20πr − 5πr2 = 0

5πr(4− r) = 0

r = 0, 4

Circular Cylinder Example

What is our domain for the radius here? 0 < r < 6

V = πr2(

10− 53

r)

=

10πr2 − 5π3

r3

dVdr

= 20πr − 5πr2 = 0

5πr(4− r) = 0

r = 0, 4

Circular Cylinder Example

What is our domain for the radius here? 0 < r < 6

V = πr2(

10− 53

r)

= 10πr2 − 5π3

r3

dVdr

= 20πr − 5πr2 = 0

5πr(4− r) = 0

r = 0, 4

Circular Cylinder Example

What is our domain for the radius here? 0 < r < 6

V = πr2(

10− 53

r)

= 10πr2 − 5π3

r3

dVdr

=

20πr − 5πr2 = 0

5πr(4− r) = 0

r = 0, 4

Circular Cylinder Example

What is our domain for the radius here? 0 < r < 6

V = πr2(

10− 53

r)

= 10πr2 − 5π3

r3

dVdr

= 20πr − 5πr2

= 0

5πr(4− r) = 0

r = 0, 4

Circular Cylinder Example

What is our domain for the radius here? 0 < r < 6

V = πr2(

10− 53

r)

= 10πr2 − 5π3

r3

dVdr

= 20πr − 5πr2 = 0

5πr(4− r) = 0

r = 0, 4

Circular Cylinder Example

What is our domain for the radius here? 0 < r < 6

V = πr2(

10− 53

r)

= 10πr2 − 5π3

r3

dVdr

= 20πr − 5πr2 = 0

5πr(4− r) = 0

r = 0, 4

Circular Cylinder Example

What is our domain for the radius here? 0 < r < 6

V = πr2(

10− 53

r)

= 10πr2 − 5π3

r3

dVdr

= 20πr − 5πr2 = 0

5πr(4− r) = 0

r = 0, 4

Circular Cylinder Example

Now what do we need to do?

d2Vdr2 = 20π − 10πr

V ′′∣∣r=0 > 0

V ′′∣∣r=4 < 0

V ′′∣∣r=6 < 0

Conclusion? We have two local maxima. Which is the right one?

Circular Cylinder Example

Now what do we need to do?

d2Vdr2 =

20π − 10πr

V ′′∣∣r=0 > 0

V ′′∣∣r=4 < 0

V ′′∣∣r=6 < 0

Conclusion? We have two local maxima. Which is the right one?

Circular Cylinder Example

Now what do we need to do?

d2Vdr2 = 20π − 10πr

V ′′∣∣r=0 > 0

V ′′∣∣r=4 < 0

V ′′∣∣r=6 < 0

Conclusion? We have two local maxima. Which is the right one?

Circular Cylinder Example

Now what do we need to do?

d2Vdr2 = 20π − 10πr

V ′′∣∣r=0

> 0

V ′′∣∣r=4 < 0

V ′′∣∣r=6 < 0

Conclusion? We have two local maxima. Which is the right one?

Circular Cylinder Example

Now what do we need to do?

d2Vdr2 = 20π − 10πr

V ′′∣∣r=0 > 0

V ′′∣∣r=4

< 0

V ′′∣∣r=6 < 0

Conclusion? We have two local maxima. Which is the right one?

Circular Cylinder Example

Now what do we need to do?

d2Vdr2 = 20π − 10πr

V ′′∣∣r=0 > 0

V ′′∣∣r=4 < 0

V ′′∣∣r=6

< 0

Conclusion? We have two local maxima. Which is the right one?

Circular Cylinder Example

Now what do we need to do?

d2Vdr2 = 20π − 10πr

V ′′∣∣r=0 > 0

V ′′∣∣r=4 < 0

V ′′∣∣r=6 < 0

Conclusion? We have two local maxima. Which is the right one?

Circular Cylinder Example

Now what do we need to do?

d2Vdr2 = 20π − 10πr

V ′′∣∣r=0 > 0

V ′′∣∣r=4 < 0

V ′′∣∣r=6 < 0

Conclusion?

We have two local maxima. Which is the right one?

Circular Cylinder Example

Now what do we need to do?

d2Vdr2 = 20π − 10πr

V ′′∣∣r=0 > 0

V ′′∣∣r=4 < 0

V ′′∣∣r=6 < 0

Conclusion? We have two local maxima. Which is the right one?

Circular Cylinder Example

V|r=6 =

360π − 360π = 0

V|r=4 = 160π − 320π3

=160π

3

So, we get the maximum volume of 160π3 when r = 4 and h = 10

3 .

Circular Cylinder Example

V|r=6 = 360π − 360π = 0

V|r=4 = 160π − 320π3

=160π

3

So, we get the maximum volume of 160π3 when r = 4 and h = 10

3 .

Circular Cylinder Example

V|r=6 = 360π − 360π = 0

V|r=4 =

160π − 320π3

=160π

3

So, we get the maximum volume of 160π3 when r = 4 and h = 10

3 .

Circular Cylinder Example

V|r=6 = 360π − 360π = 0

V|r=4 = 160π − 320π3

=160π

3

So, we get the maximum volume of 160π3 when r = 4 and h = 10

3 .

Circular Cylinder Example

V|r=6 = 360π − 360π = 0

V|r=4 = 160π − 320π3

=160π

3

So, we get the maximum volume of 160π3 when r = 4 and h = 10

3 .

Shipping Example

ExampleBoise, Idaho is about 300 miles inland from the nearest point on thePacific Coast. San Diego is about 1000 miles south of that point downthe coast. Assuming the coast is a straight line going north to south, Cis the point on the coast directly west of Boise. It costs $.02 per mileto transport a ton of potatoes and $.01 to transport them by ship. TheIdaho Potato Company wants to find a point P, on the Pacific Coast,to which it should truck the potatoes before loading them aboard acargo ship in order to minimize cost.

Let’s start with a visual.

Shipping Example

ExampleBoise, Idaho is about 300 miles inland from the nearest point on thePacific Coast. San Diego is about 1000 miles south of that point downthe coast. Assuming the coast is a straight line going north to south, Cis the point on the coast directly west of Boise. It costs $.02 per mileto transport a ton of potatoes and $.01 to transport them by ship. TheIdaho Potato Company wants to find a point P, on the Pacific Coast,to which it should truck the potatoes before loading them aboard acargo ship in order to minimize cost.

Let’s start with a visual.

Shipping Example

300 mi

1000 mi

BC

SD

Px

1000− x√

x2 + 90000

Shipping Example

300 mi

1000 mi

BC

SD

P

x

1000− x√

x2 + 90000

Shipping Example

300 mi

1000 mi

BC

SD

P

x

1000− x√

x2 + 90000

Shipping Example

300 mi

1000 mi

BC

SD

Px

1000− x√

x2 + 90000

Shipping Example

300 mi

1000 mi

BC

SD

Px

1000− x

√x2 + 90000

Shipping Example

300 mi

1000 mi

BC

SD

Px

1000− x

√x2 + 90000

Shipping Example

300 mi

1000 mi

BC

SD

Px

1000− x√

x2 + 90000

Shipping Example

What is the function we want to optimize?

D =√

x2 + 90000 + (1000− x)

C = 2√

x2 + 90000 + 1(1000− x)

Notice we didn’t use $.02 and $.01. This is because those decimalswould be harder to work with and the fact that the only thing thatreally matters is that it costs twice as much over land as it does byboat.

Shipping Example

What is the function we want to optimize?

D =√

x2 + 90000 + (1000− x)

C = 2√

x2 + 90000 + 1(1000− x)

Notice we didn’t use $.02 and $.01. This is because those decimalswould be harder to work with and the fact that the only thing thatreally matters is that it costs twice as much over land as it does byboat.

Shipping Example

What is the function we want to optimize?

D =√

x2 + 90000 + (1000− x)

C = 2√

x2 + 90000 + 1(1000− x)

Notice we didn’t use $.02 and $.01. This is because those decimalswould be harder to work with and the fact that the only thing thatreally matters is that it costs twice as much over land as it does byboat.

Shipping Example

What is the function we want to optimize?

D =√

x2 + 90000 + (1000− x)

C = 2√

x2 + 90000 + 1(1000− x)

Notice we didn’t use $.02 and $.01. This is because those decimalswould be harder to work with and the fact that the only thing thatreally matters is that it costs twice as much over land as it does byboat.

Shipping Example

C′ =

2x√x2 + 90000

− 1 = 0

2x√x2 + 90000

= 1

4x2

x2 + 90000= 1

4x2 = x2 + 90000

3x2 = 90000

x = ±173.21

The negative value is impossible, so we need only consider outdomain’s endpoints and the critical point.

Shipping Example

C′ =2x√

x2 + 90000− 1

= 0

2x√x2 + 90000

= 1

4x2

x2 + 90000= 1

4x2 = x2 + 90000

3x2 = 90000

x = ±173.21

The negative value is impossible, so we need only consider outdomain’s endpoints and the critical point.

Shipping Example

C′ =2x√

x2 + 90000− 1 = 0

2x√x2 + 90000

= 1

4x2

x2 + 90000= 1

4x2 = x2 + 90000

3x2 = 90000

x = ±173.21

The negative value is impossible, so we need only consider outdomain’s endpoints and the critical point.

Shipping Example

C′ =2x√

x2 + 90000− 1 = 0

2x√x2 + 90000

= 1

4x2

x2 + 90000= 1

4x2 = x2 + 90000

3x2 = 90000

x = ±173.21

The negative value is impossible, so we need only consider outdomain’s endpoints and the critical point.

Shipping Example

C′ =2x√

x2 + 90000− 1 = 0

2x√x2 + 90000

= 1

4x2

x2 + 90000= 1

4x2 = x2 + 90000

3x2 = 90000

x = ±173.21

The negative value is impossible, so we need only consider outdomain’s endpoints and the critical point.

Shipping Example

C′ =2x√

x2 + 90000− 1 = 0

2x√x2 + 90000

= 1

4x2

x2 + 90000= 1

4x2 = x2 + 90000

3x2 = 90000

x = ±173.21

The negative value is impossible, so we need only consider outdomain’s endpoints and the critical point.

Shipping Example

C′ =2x√

x2 + 90000− 1 = 0

2x√x2 + 90000

= 1

4x2

x2 + 90000= 1

4x2 = x2 + 90000

3x2 = 90000

x = ±173.21

The negative value is impossible, so we need only consider outdomain’s endpoints and the critical point.

Shipping Example

C′ =2x√

x2 + 90000− 1 = 0

2x√x2 + 90000

= 1

4x2

x2 + 90000= 1

4x2 = x2 + 90000

3x2 = 90000

x = ±173.21

The negative value is impossible, so we need only consider outdomain’s endpoints and the critical point.

Shipping Example

C′ =2x√

x2 + 90000− 1 = 0

2x√x2 + 90000

= 1

4x2

x2 + 90000= 1

4x2 = x2 + 90000

3x2 = 90000

x = ±173.21

The negative value is impossible, so we need only consider outdomain’s endpoints and the critical point.

Shipping Example

What is the domain?

D(C) : 0 ≤ x ≤ 1000

What happens if we use x = 0? We are not cutting any travel out sincethat would entail the whole distance over land and the whole distanceover the ocean.

What happens if we use x = 1000? We are covering the entiredistance over land, which is the more expensive way to go and couldnot possibly minimize cost. So the endpoints won’t work.

Are we done?

We need to make sure the critical point is a minimum.

Shipping Example

What is the domain? D(C) : 0 ≤ x ≤ 1000

What happens if we use x = 0? We are not cutting any travel out sincethat would entail the whole distance over land and the whole distanceover the ocean.

What happens if we use x = 1000? We are covering the entiredistance over land, which is the more expensive way to go and couldnot possibly minimize cost. So the endpoints won’t work.

Are we done?

We need to make sure the critical point is a minimum.

Shipping Example

What is the domain? D(C) : 0 ≤ x ≤ 1000

What happens if we use x = 0?

We are not cutting any travel out sincethat would entail the whole distance over land and the whole distanceover the ocean.

What happens if we use x = 1000? We are covering the entiredistance over land, which is the more expensive way to go and couldnot possibly minimize cost. So the endpoints won’t work.

Are we done?

We need to make sure the critical point is a minimum.

Shipping Example

What is the domain? D(C) : 0 ≤ x ≤ 1000

What happens if we use x = 0? We are not cutting any travel out sincethat would entail the whole distance over land and the whole distanceover the ocean.

What happens if we use x = 1000? We are covering the entiredistance over land, which is the more expensive way to go and couldnot possibly minimize cost. So the endpoints won’t work.

Are we done?

We need to make sure the critical point is a minimum.

Shipping Example

What is the domain? D(C) : 0 ≤ x ≤ 1000

What happens if we use x = 0? We are not cutting any travel out sincethat would entail the whole distance over land and the whole distanceover the ocean.

What happens if we use x = 1000?

We are covering the entiredistance over land, which is the more expensive way to go and couldnot possibly minimize cost. So the endpoints won’t work.

Are we done?

We need to make sure the critical point is a minimum.

Shipping Example

What is the domain? D(C) : 0 ≤ x ≤ 1000

What happens if we use x = 0? We are not cutting any travel out sincethat would entail the whole distance over land and the whole distanceover the ocean.

What happens if we use x = 1000? We are covering the entiredistance over land, which is the more expensive way to go and couldnot possibly minimize cost. So the endpoints won’t work.

Are we done?

We need to make sure the critical point is a minimum.

Shipping Example

What is the domain? D(C) : 0 ≤ x ≤ 1000

What happens if we use x = 0? We are not cutting any travel out sincethat would entail the whole distance over land and the whole distanceover the ocean.

What happens if we use x = 1000? We are covering the entiredistance over land, which is the more expensive way to go and couldnot possibly minimize cost. So the endpoints won’t work.

Are we done?

We need to make sure the critical point is a minimum.

Shipping Example

What is the domain? D(C) : 0 ≤ x ≤ 1000

What happens if we use x = 0? We are not cutting any travel out sincethat would entail the whole distance over land and the whole distanceover the ocean.

What happens if we use x = 1000? We are covering the entiredistance over land, which is the more expensive way to go and couldnot possibly minimize cost. So the endpoints won’t work.

Are we done?

We need to make sure the critical point is a minimum.

Shipping Example

C′′ =

2√

x2 + 90000− 4x2√

x2+90000

x2 + 90000

=−2x2 + 180000

(x2 + 90000)32

C′′∣∣x=173.21 > 0

Conclusion? This point does give a local minimum. Therefore, thepoint P where the potatoes should be loaded onto a boat is at 173.21miles due south of point C.

Shipping Example

C′′ =2√

x2 + 90000− 4x2√

x2+90000

x2 + 90000

=−2x2 + 180000

(x2 + 90000)32

C′′∣∣x=173.21 > 0

Conclusion? This point does give a local minimum. Therefore, thepoint P where the potatoes should be loaded onto a boat is at 173.21miles due south of point C.

Shipping Example

C′′ =2√

x2 + 90000− 4x2√

x2+90000

x2 + 90000

=−2x2 + 180000

(x2 + 90000)32

C′′∣∣x=173.21 > 0

Conclusion? This point does give a local minimum. Therefore, thepoint P where the potatoes should be loaded onto a boat is at 173.21miles due south of point C.

Shipping Example

C′′ =2√

x2 + 90000− 4x2√

x2+90000

x2 + 90000

=−2x2 + 180000

(x2 + 90000)32

C′′∣∣x=173.21

> 0

Conclusion? This point does give a local minimum. Therefore, thepoint P where the potatoes should be loaded onto a boat is at 173.21miles due south of point C.

Shipping Example

C′′ =2√

x2 + 90000− 4x2√

x2+90000

x2 + 90000

=−2x2 + 180000

(x2 + 90000)32

C′′∣∣x=173.21 > 0

Conclusion? This point does give a local minimum. Therefore, thepoint P where the potatoes should be loaded onto a boat is at 173.21miles due south of point C.

Shipping Example

C′′ =2√

x2 + 90000− 4x2√

x2+90000

x2 + 90000

=−2x2 + 180000

(x2 + 90000)32

C′′∣∣x=173.21 > 0

Conclusion?

This point does give a local minimum. Therefore, thepoint P where the potatoes should be loaded onto a boat is at 173.21miles due south of point C.

Shipping Example

C′′ =2√

x2 + 90000− 4x2√

x2+90000

x2 + 90000

=−2x2 + 180000

(x2 + 90000)32

C′′∣∣x=173.21 > 0

Conclusion? This point does give a local minimum. Therefore, thepoint P where the potatoes should be loaded onto a boat is at 173.21miles due south of point C.